Calculus Examples

$y (t) = \frac{c}{b - a} \cdot (e^{- a t} - e^{- b t})$

Step 1

Differentiate.

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Step 1.1

Since $\frac{c}{b - a}$ is constant with respect to $t$ , the derivative of $\frac{c}{b - a} (e^{- a t} - e^{- b t})$ with respect to $t$ is $\frac{c}{b - a} \frac{d}{d t} [e^{- a t} - e^{- b t}]$ .

$\frac{c}{b - a} \frac{d}{d t} [e^{- a t} - e^{- b t}]$

Step 1.2

By the Sum Rule, the derivative of $e^{- a t} - e^{- b t}$ with respect to $t$ is $\frac{d}{d t} [e^{- a t}] + \frac{d}{d t} [- e^{- b t}]$ .

$\frac{c}{b - a} (\frac{d}{d t} [e^{- a t}] + \frac{d}{d t} [- e^{- b t}])$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - a t$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $- a t$ .

$\frac{c}{b - a} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d t} [- a t] + \frac{d}{d t} [- e^{- b t}])$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$\frac{c}{b - a} (e^{u_{1}} \frac{d}{d t} [- a t] + \frac{d}{d t} [- e^{- b t}])$

Step 2.3

Replace all occurrences of $u_{1}$ with $- a t$ .

$\frac{c}{b - a} (e^{- a t} \frac{d}{d t} [- a t] + \frac{d}{d t} [- e^{- b t}])$

Step 3

Differentiate.

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Step 3.1

Since $- a$ is constant with respect to $t$ , the derivative of $- a t$ with respect to $t$ is $- a \frac{d}{d t} [t]$ .

$\frac{c}{b - a} (e^{- a t} (- a \frac{d}{d t} [t]) + \frac{d}{d t} [- e^{- b t}])$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{c}{b - a} (e^{- a t} (- a \cdot 1) + \frac{d}{d t} [- e^{- b t}])$

Step 3.3

Multiply $- 1$ by $1$ .

$\frac{c}{b - a} (e^{- a t} (- a) + \frac{d}{d t} [- e^{- b t}])$

Step 3.4

Since $- 1$ is constant with respect to $t$ , the derivative of $- e^{- b t}$ with respect to $t$ is $- \frac{d}{d t} [e^{- b t}]$ .

$\frac{c}{b - a} (e^{- a t} (- a) - \frac{d}{d t} [e^{- b t}])$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - b t$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $- b t$ .

$\frac{c}{b - a} (e^{- a t} (- a) - (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [- b t]))$

Step 4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$\frac{c}{b - a} (e^{- a t} (- a) - (e^{u_{2}} \frac{d}{d t} [- b t]))$

Step 4.3

Replace all occurrences of $u_{2}$ with $- b t$ .

$\frac{c}{b - a} (e^{- a t} (- a) - (e^{- b t} \frac{d}{d t} [- b t]))$

Step 5

Differentiate.

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Step 5.1

Since $- b$ is constant with respect to $t$ , the derivative of $- b t$ with respect to $t$ is $- b \frac{d}{d t} [t]$ .

$\frac{c}{b - a} (e^{- a t} (- a) - e^{- b t} (- b \frac{d}{d t} [t]))$

Step 5.2

Multiply.

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Step 5.2.1

Multiply $- 1$ by $- 1$ .

$\frac{c}{b - a} (e^{- a t} (- a) + 1 e^{- b t} (b \frac{d}{d t} [t]))$

Step 5.2.2

Multiply $e^{- b t}$ by $1$ .

$\frac{c}{b - a} (e^{- a t} (- a) + e^{- b t} (b \frac{d}{d t} [t]))$

Step 5.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{c}{b - a} (e^{- a t} (- a) + e^{- b t} (b \cdot 1))$

Step 5.4

Multiply $b$ by $1$ .

$\frac{c}{b - a} (e^{- a t} (- a) + e^{- b t} b)$

Step 6

Simplify.

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Step 6.1

Reorder the factors of $\frac{c}{b - a} (e^{- a t} (- a) + e^{- b t} b)$ .

$(e^{- a t} (- a) + e^{- b t} b) \frac{c}{b - a}$

Step 6.2

Rewrite using the commutative property of multiplication.

$(- e^{- a t} a + e^{- b t} b) \frac{c}{b - a}$

Step 6.3

Multiply $- e^{- a t} a + e^{- b t} b$ by $\frac{c}{b - a}$ .

$\frac{(- e^{- a t} a + e^{- b t} b) c}{b - a}$

Step 6.4

Factor $- 1$ out of $- e^{- a t} a$ .

$\frac{(- (e^{- a t} a) + e^{- b t} b) c}{b - a}$

Step 6.5

Factor $- 1$ out of $e^{- b t} b$ .

$\frac{(- (e^{- a t} a) - (- e^{- b t} b)) c}{b - a}$

Step 6.6

Factor $- 1$ out of $- (e^{- a t} a) - (- e^{- b t} b)$ .

$\frac{- (e^{- a t} a - e^{- b t} b) c}{b - a}$

Step 6.7

Rewrite $- (e^{- a t} a - e^{- b t} b)$ as $- 1 (e^{- a t} a - e^{- b t} b)$ .

$\frac{- 1 (e^{- a t} a - e^{- b t} b) c}{b - a}$

Step 6.8

Move the negative in front of the fraction.

$- \frac{(e^{- a t} a - e^{- b t} b) c}{b - a}$

Step 6.9

Reorder factors in $- \frac{(e^{- a t} a - e^{- b t} b) c}{b - a}$ .

$- \frac{c (a e^{- a t} - b e^{- b t})}{b - a}$